In an arithmetic progression, if S_40=1030 and S_12=57, then S_30-S_10 is equal to :

Question

In an arithmetic progression, if ${S}_{40}=1030$ and ${S}_{12}=57$, then ${S}_{30}-{S}_{10}$ is equal to :

Accepted Answer

Let a \& d are first term and common diff of an AP.

$$\begin{aligned}
& \mathrm{S}_{40}=\frac{40}{2}[2 \mathrm{a}+39 \mathrm{~d}]=1030 \quad\text{..... (1)}\
& \mathrm{~S}_{12}=\frac{12}{2}[2 \mathrm{a}+11 \mathrm{~d}]=57 \quad\text{..... (2)}
\end{aligned}$$

by (1) \& (2)

$$\begin{aligned}
& \mathrm{a}=-\frac{7}{2} \quad \mathrm{~d}=\frac{3}{2} \
& \therefore \mathrm{~S}_{30}-\mathrm{S}_{10}=\frac{30}{2}[2 \mathrm{a}+29 \mathrm{~d}]-\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}] \
& =20 \mathrm{a}-390 \mathrm{~d} \
& =515
\end{aligned}$$

Solutions · 1